Redacted

I'm not sure I understand. Are you claiming that $n^5-n^2$ is infinitely often a perfect square? Or what excactly is the relation between the elliptic curve and the original problem?

Redacted, nvm i was brain dead when I wrote this.

I understand that $y^3=x^2+1$ has infinitely many rational solutions. But I still don't understand how this is related to the problem. Again, if you are claiming that $n^5-n^2$ is infinitely often a perfect square, this is just plain wrong.

Some possible approach: $n^5+2n+1=a^2+b^2,~~ (a,b)=1,~~~ n=2k+1$ $\Longrightarrow n^5+2n+1=(2k+1)^5+2(2k+1)+1=a^2+b^2$ all odd prime divisors of $~~a^2+b^2~~$ have to be congruent to $1 \bmod {4}$ $\Longrightarrow \frac{(2k+1)^5+2(2k+1)+1}{2}=\frac{32k^5+80k^4+80k^3+40k^2+14k+4}{2}=$ $16k^5+40k^4+40k^3+20k^2+7k+2=4M+1$ $\Longrightarrow 7k+2=1 \bmod {4}\Longrightarrow k=4t+1\Longrightarrow n=2k+1=2(4t+1)+1=8t+3$ $t=0 \longrightarrow n=3\longrightarrow \frac{3^5+6+1}{2}=125=2^2+11^2, ~~(2,11)=1.$ $t=1 \longrightarrow n=11\longrightarrow \frac{11^5+22+1}{2}=80537=116^2+259^2,~~ (116,259)=1.$ $t=3 \longrightarrow n=27\longrightarrow \frac{27^5+54+1}{2}=7174481=1779^2+2020^2, ~~(1779,2020)=1$. $t=2, 5 \longrightarrow n=19, 43\longrightarrow $ do not work (presence of two $"4s+3"$ prime factors) Although it cannot be taken for granted that an even number of $"4s+3"$ prime factors would not turn up, most cases work.

Now that almost a year has passed, here's the official solution: Write $4(n+2)(n^5+2n+1)=(2n^3+2n^2-n+1)^2+(3n^2+22n+7)$. Since $3n^2+22n+7$ is a square for infinitely many odd integers, we are done. (Using the theory of Pell equations we just need to check there is indeed a solution, and hopefully $n=3$ works. A little more work is needed to verify that $(2n^3+2n^2-n+1, 3n^2+22n+7)=1$ for these values.) $\square$ The motivation behind this approach is that we want to make one term as a square of a polynomial, and the rest with low degree terms. The first thing that comes to mind to eliminate $n^5$ is to let $n$ be a square itself, but sadly for that case $2n+1\equiv 3\pmod{4}$. Twisting this and letting $n$ be a monic quadratic fails because the best you can get for the remaining term is of degree $4$. This gives insight that for this idea to succeed, one need to make a polynomial of degree at most $6$. Now that there is only degree $1$ that we can add from the starting polynomial, we figure that multiplying a linear term is our best attempt. (This is compatible to the fact that if $AB$ is a sum of two coprime squares, then so are $A$ and $B$.) On a side note, there was another beautiful solution by one of the students that exploits the fact that the linear term "looks" like it comes from $n^2+2n+1$. The sketch of the proof is to write $n^5+2n+1=n^2(n^3-a^2n^2-2an-2a-1)+(an^2+n+1)^2$, and find infinitely many $(n,a)$ such $n^3-a^2n^2-2an-2a-1$ is a square.

I remember that some problems from the AoPS user kaede has the same solution. Maybe the proposer was aware of it?

Show that there are infinitely many positive odd integers $n$ such that $n^5+2n+1$ is expressible as a sum of squares of two coprime integers: Doesn’t $n =(2k+1)^2 -2$ satisfy this? Then $$n^5+2n+1=[((2k+1)^2 -2)^2(2k+1)]^2 + 1^2$$ Or am I making some silly mistake?